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  3. one equal to two ?

one equal to two ?

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  • P PIEBALDconsult

    Abbott and Costello said it better.

    M Offline
    M Offline
    Mark_Wallace
    wrote on last edited by
    #55

    PIEBALDconsult wrote:

    Abbott and Costello_, who were on first,_ said it better.

    It was missing something.

    I wanna be a eunuchs developer! Pass me a bread knife!

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    • T tayoufabrice

      Let read this : 1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2 Where is the error ?

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      L Offline
      Lost User
      wrote on last edited by
      #56

      welcome to the lounge. I know this isn't quite your first post - but nearly. And I would like to apologise for the negativity your post received. For someone who hadn't seen that 'proof' before it may have been interesting - as you can see, not only have the majority here seen it (more than once!) but they like to stuff it down your throat - whether to big-note themselves or simply in an attempt to belittle you we cannot tell. They should be ashamed. Merry Xmas

      PooperPig - Coming Soon

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      • T tayoufabrice

        Let read this : 1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2 Where is the error ?

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        M Offline
        Mitchell J
        wrote on last edited by
        #57

        How about this... ;P Start with this: 1/9 = 1/9 Then convert one side to decimal equivalent (which is infinitely recurring) 1/9 = 0.11111111111111111111111111111111111111111111111111111111...(etc etc) Then multiply both sides by nine 1 = 0.9999999999999999999999999999999999999999999999999999999999...(etc etc) Therefore, 1 is equal to 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...(where the 9's are in infinite recursion). And yes, this actually is mathematically correct.

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        • L Lost User

          welcome to the lounge. I know this isn't quite your first post - but nearly. And I would like to apologise for the negativity your post received. For someone who hadn't seen that 'proof' before it may have been interesting - as you can see, not only have the majority here seen it (more than once!) but they like to stuff it down your throat - whether to big-note themselves or simply in an attempt to belittle you we cannot tell. They should be ashamed. Merry Xmas

          PooperPig - Coming Soon

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          T Offline
          tayoufabrice
          wrote on last edited by
          #58

          Thank you Max and happy new Xear

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          • M Mitchell J

            How about this... ;P Start with this: 1/9 = 1/9 Then convert one side to decimal equivalent (which is infinitely recurring) 1/9 = 0.11111111111111111111111111111111111111111111111111111111...(etc etc) Then multiply both sides by nine 1 = 0.9999999999999999999999999999999999999999999999999999999999...(etc etc) Therefore, 1 is equal to 0.9999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999...(where the 9's are in infinite recursion). And yes, this actually is mathematically correct.

            T Offline
            T Offline
            tayoufabrice
            wrote on last edited by
            #59

            I agree but here 1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9 I could write 1/9~= 0.111111111111111111111111...... then 1 ~= 0.9999999999999999999999999...... ??

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            • T tayoufabrice

              I agree but here 1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9 I could write 1/9~= 0.111111111111111111111111...... then 1 ~= 0.9999999999999999999999999...... ??

              M Offline
              M Offline
              Mitchell J
              wrote on last edited by
              #60

              tayoufabrice wrote:

              1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9

              Wish I could agree, but I can't... read all about it[^] :-D Even google 0.999999999999999 = 1[^] if you're still unconvinced.

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              • M Mitchell J

                tayoufabrice wrote:

                1/9 = 0.111111111111111111111111...... is not really true ; we lost 0.000000000000000000000.......9

                Wish I could agree, but I can't... read all about it[^] :-D Even google 0.999999999999999 = 1[^] if you're still unconvinced.

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                tayoufabrice
                wrote on last edited by
                #61

                Ah là là :laugh: Mathematics !! (French laughing)

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                • T tayoufabrice

                  Let read this : 1=1 a=a a²=a² a²-a²=a²-a² a(a-a)=(a+a)(a-a) a=a+a a(1)=a(1+1) 1=1+1 1=2 Where is the error ?

                  D Offline
                  D Offline
                  Daniel Pfeffer
                  wrote on last edited by
                  #62

                  a(a-a) = (a+a)(a-a) // divide by (a-a), i.e. divide by 0 a = a+a Division by zero is a no-no because it can lead to "impossible" results like the above.

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                  • D Daniel Pfeffer

                    a(a-a) = (a+a)(a-a) // divide by (a-a), i.e. divide by 0 a = a+a Division by zero is a no-no because it can lead to "impossible" results like the above.

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                    tayoufabrice
                    wrote on last edited by
                    #63

                    I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded) Now ??

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                    • T tayoufabrice

                      I could fix the post as : Given a C ]--;0[ U ]0;++[ (meaning 0 excluded) Now ??

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                      Daniel Pfeffer
                      wrote on last edited by
                      #64

                      The value of a is irrelevant; a - a == 0, and factoring out a - a is division by 0, which is forbidden. I am not a mathematician, so I don't know if it is possible to create a self-consistent arithmetic in which division by 0 does not result in nonsensical results. All I know is that in the arithmetic I learnt in school it is forbidden.

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                      • D Daniel Pfeffer

                        The value of a is irrelevant; a - a == 0, and factoring out a - a is division by 0, which is forbidden. I am not a mathematician, so I don't know if it is possible to create a self-consistent arithmetic in which division by 0 does not result in nonsensical results. All I know is that in the arithmetic I learnt in school it is forbidden.

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                        tayoufabrice
                        wrote on last edited by
                        #65

                        Sure ! :laugh: number can never be divided by zero 0 ; even 0/0 :confused: It is the real error of my process

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